This question is from my assignment in linear algebra and I am not able to make any significant progress on this.

Question: Let $A$ be a $5\times 4$ matrix with real entries such that $Ax=0$ iff $x=0$ where $x$ is a $4\times 1$ vector and $0$ is a null vector. Then the rank of $A$ is ?

I am sorry but I don't have any intuition regarding what should be done in this question.

I have been following the textbook Linear Algebra by Hoffman and Kunze.

Kindly help.


1 Answer 1


Forget matrices. Matrices are only the computional aspect of linear algebra, which in small dimensions up to $3$ is elementary geometry.

For example, see $A$ as the matrix of a linear map $$u:\mathbb R^4\to \mathbb R^5$$ in canonical basis.

Then you have $$\dim \ker u+\dim Im (u)=4$$ because of $$\begin{matrix}\mathbb R^4 & \overset{u}{\longrightarrow} & \mathbb R^5 \\\downarrow & & \uparrow \\ \mathbb R^4/\ker u & \overset{\approx }{\longrightarrow} & Im(u) \\\end{matrix}$$

Your hypothesis translates into the fact that $\ker u=\{0\}$

So, $$rank A:=rank (u):=\dim Im(u)=4.$$

  • $\begingroup$ Why not write $A$ everywhere you write $u$? $\endgroup$
    – Elmex80s
    Apr 10 at 8:15
  • $\begingroup$ To avoid a notation clash. $A=Mat(u; e,f)$, where $e,f$ are basis. It depends on $e$ and $f$ chosen. But rank is intrinsically linked to $u$. $\endgroup$ Apr 10 at 8:17
  • $\begingroup$ In other words, $u$ is not $A$. A morphism between vector spaces $V$ and $W$ is not the matrix that represents it in basis of $V$ and $W$. It's just a useful calculation tool for $u$. $\endgroup$ Apr 10 at 8:24

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